Differentiable function with no second derivative at $0$?
What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?
calculus analysis examples-counterexamples
edited Dec 6 at 16:59
Andrews
327217
asked Dec 6 at 16:49
Y.Z.
243
add a comment |
What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?
calculus analysis examples-counterexamples
edited Dec 6 at 16:59
Andrews
327217
asked Dec 6 at 16:49
Y.Z.
243
3
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
– Michael Hoppe
Dec 6 at 16:57
@Eric Maybe only integral of absolute value?
– Юрій Ярош
Dec 6 at 17:01
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
– Eric
Dec 6 at 17:04
add a comment |
What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?
calculus analysis examples-counterexamples
edited Dec 6 at 16:59
Andrews
327217
asked Dec 6 at 16:49
Y.Z.
243
What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?
calculus analysis examples-counterexamples
calculus analysis examples-counterexamples
edited Dec 6 at 16:59
Andrews
327217
asked Dec 6 at 16:49
Y.Z.
243
edited Dec 6 at 16:59
Andrews
327217
asked Dec 6 at 16:49
Y.Z.
243
edited Dec 6 at 16:59
Andrews
327217
edited Dec 6 at 16:59
Andrews
327217
edited Dec 6 at 16:59
Andrews
327217
327217
asked Dec 6 at 16:49
Y.Z.
243
asked Dec 6 at 16:49
Y.Z.
243
asked Dec 6 at 16:49
Y.Z.
243
243
3
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
– Michael Hoppe
Dec 6 at 16:57
@Eric Maybe only integral of absolute value?
– Юрій Ярош
Dec 6 at 17:01
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
– Eric
Dec 6 at 17:04
add a comment |
3
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
– Michael Hoppe
Dec 6 at 16:57
@Eric Maybe only integral of absolute value?
– Юрій Ярош
Dec 6 at 17:01
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
– Eric
Dec 6 at 17:04
3
3
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
– Michael Hoppe
Dec 6 at 16:57
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
– Michael Hoppe
Dec 6 at 16:57
@Eric Maybe only integral of absolute value?
– Юрій Ярош
Dec 6 at 17:01
@Eric Maybe only integral of absolute value?
– Юрій Ярош
Dec 6 at 17:01
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
– Eric
Dec 6 at 17:04
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
– Eric
Dec 6 at 17:04
add a comment |
1 Answer
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So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 at 16:59
Andrei
10.8k21025
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
– zhw.
Dec 6 at 17:32
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
– Andrei
Dec 6 at 19:22
add a comment |
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So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 at 16:59
Andrei
10.8k21025
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
– zhw.
Dec 6 at 17:32
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
– Andrei
Dec 6 at 19:22
add a comment |
So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 at 16:59
Andrei
10.8k21025
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
– zhw.
Dec 6 at 17:32
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
– Andrei
Dec 6 at 19:22
add a comment |
So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 at 16:59
Andrei
10.8k21025
So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$
answered Dec 6 at 16:59
Andrei
10.8k21025
answered Dec 6 at 16:59
Andrei
10.8k21025
answered Dec 6 at 16:59
Andrei
10.8k21025
answered Dec 6 at 16:59
Andrei
10.8k21025
10.8k21025
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
– zhw.
Dec 6 at 17:32
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
– Andrei
Dec 6 at 19:22
add a comment |
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
– zhw.
Dec 6 at 17:32
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
– Andrei
Dec 6 at 19:22
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
– zhw.
Dec 6 at 17:32
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
– zhw.
Dec 6 at 17:32
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
– Andrei
Dec 6 at 19:22
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
– Andrei
Dec 6 at 19:22
add a comment |
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3
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
– Michael Hoppe
Dec 6 at 16:57
@Eric Maybe only integral of absolute value?
– Юрій Ярош
Dec 6 at 17:01
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
– Eric
Dec 6 at 17:04